Show that in a parallel plate capacitor the stored electric energy is Ue=C*(V^2)/2 where C is the capacity and V the voltage. (Hint: Remember that in a capacitor Q=C*V)

Question

Thanks.

Answer 1

Your first step, is that you consider the electric field across the plates of the capacitor. When you take any small charge = dq from one plate of the capacitor to another, work will be done due to the presence of this electric field. That work is essentially the energy that the capacitor stores.

For a small charge dq, the work done in this process = dW = V*dq

Because we know that work is equal to the charge multiplied by the potential difference

Now, plug in V = Q/C, which is the standard equation for a capacitor.

You get dW = (1/C)*(Q*dq)

Integrate both sides, and you'll get your term for energy stored in a capacitor.

Answer 2

or say power = V i = V dQ/dt
but
Q = V C
so
V i = (1/C) Q dQ/dt (same as Arora did)
energy = power *time = (1/C) Q^2/2 = C V^2/2